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Thread: inequality

  1. #1
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    inequality

    when i have an inequality like this ,

    $\displaystyle 0<\sqrt{13}-3<1$

    If i were to find $\displaystyle (\sqrt{13}-3)^4$ , can i say that

    $\displaystyle 0^4<(\sqrt{13}-3)^4<1^4 = 0<(\sqrt{13}-3)^4<1$ ??

    And if the value at the side are negative , for instance

    $\displaystyle
    -5<x<-1
    $
    If i were find range of $\displaystyle x^2$ , am i right in saying .

    $\displaystyle 25>x^2>1$

    Rearrange $\displaystyle 1<x^2<25$
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  2. #2
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    Yes, you can
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  3. #3
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    Quote Originally Posted by songoku View Post
    Yes, you can

    THanks .

    How about only one of its side is negative ?

    $\displaystyle -5<x<2 $

    TO find $\displaystyle x^2 $, do i do that this way :

    $\displaystyle 25>x^2>4 $

    Then rearrange $\displaystyle 4<x^2<25$
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  4. #4
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    Quote Originally Posted by thereddevils View Post
    THanks .

    How about only one of its side is negative ?

    $\displaystyle -5<x<2 $

    TO find $\displaystyle x^2 $, do i do that this way :

    $\displaystyle 25>x^2>4 $

    Then rearrange $\displaystyle 4<x^2<25$
    x = 0 satisfies $\displaystyle -5<x<2 $. Does x = 0 satisfy $\displaystyle 4<x^2<25$ ....?
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    x = 0 satisfies $\displaystyle -5<x<2 $. Does x = 0 satisfy $\displaystyle 4<x^2<25$ ....?

    THanks , yeah it doesn't . But i still cant figure out how to find the range of x^2 .
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  6. #6
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    Moral: do that as long as all pieces are positive.
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  7. #7
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    Quote Originally Posted by thereddevils View Post
    THanks , yeah it doesn't . But i still cant figure out how to find the range of x^2 .
    The range of $\displaystyle f(x)= x^2$ is the set of all non-negative numbers.
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  8. #8
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    Quote Originally Posted by thereddevils View Post
    How about only one of its side is negative ?

    $\displaystyle -5<x<2 $

    TO find $\displaystyle x^2 $, do i do that this way :

    math]25>x^2>4 [/tex]

    Then rearrange $\displaystyle 4<x^2<25$
    Quote Originally Posted by mr fantastic View Post
    x = 0 satisfies $\displaystyle -5<x<2 $. Does x = 0 satisfy $\displaystyle 4<x^2<25$ ....?
    Quote Originally Posted by thereddevils View Post
    THanks , yeah it doesn't . But i still cant figure out how to find the range of x^2 .
    Draw a graph of y = x^2 for -5 < x < 2. Now look at the range of values of x^2.
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